CIAL Interview Questions, Process, and Tips

21

C

O

200

F+V-E =2;
F= faces;V= vertices;E = number of edges

Sunil Shetty

( b ) ANGEME

A

(1/2)*x*y

Hard work would mean spending long hours to complete my work without any shortcuts. It definitely ensures results but the process is long and stressful. Smart work would be aiming for the same results but with planning and prioritization of tasks.

@ Aprameya Shyam
length=x does not imply breadth=36-x
2x+4y=36.
3 possible answers are
(2,8) -> 16sqcm.
(4,7) -> 28sqcm.
(6,6) -> 36sqcm

let x be lenghth of candles.

thicker one—- in 1 hr 1/6th will go.
thinner one— in 1 hr 1/4th will go.

let he lit for n hrs.

so in n hrs thicker one goes n/6.
and thinner one goes n/4.

so remainini length are x-n/6 and x-n/4.so x-n/6=2(x-n/4)
implie n=3x. if x=1 n=3hrs.

Either 4 or (-)4

To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.

The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

There are 24 primes in S, so the product of S is:

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97

We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.

5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.

25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.

Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros

Let the principal be P and rate of interest be R%.
Therefore Required ratio

=P×R×6/100/P×R×9/100
=6PR/9PR
=6/9
=2:3

7

300%

Son in law

( d ) 8404